1/1*3+1/3*5+1/5*7+......+1/17*19+1/19*21

a1=1/(1*3)
a2=1/(3*5)
a3=1/(5*7)
...
那么
an=1/[(2k-1)*(2k+1)]=(1/2)[1/(2k-1)-1/(2k+1)] (k∈N)
sn=a1+a2+...+an=1/(1*3)+1/(3*5)+1/(5*7)+...+1/(17*19)+1/(19*21)
=(1/2)(1-1/3+1/3-1/5+1/5-1/7+...+1/17-1/19+1/19-1/21)
=(1/2)(1-1/21)
=10/21
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另:形如"1/(a*b)" 变为 "(1/a)-(1/b) "的方法:

(1/a)-(1/b)=(b-a)/(a*b)=(b-a)*[1/(a*b)]

=(1-1/3+1/3-1/5+1/5-1/7+……+1/17-1/19+1/19-1/21)/2
=(1-1/21)/2
=(20/21)/2
=10/21
望采纳~